55rd Ukrainian National Mathematical Olympiad - Third Round

Since $B{C}_{1}H{A}_1$ and $BANC$ are cyclic quadrilaterals, it follows that $\angle A_{1}H{C}_1=180^{\circ }-\angle B$ and $\angle ANC=180^{\circ }-\angle B$. This implies $\angle A_{1}H{C}_1=\angle ANC$. But $\angle ANC=\angle A_{1}H{C}_1$ (as vertical angles) and $\angle AKC=\angle ANC$ (as opposite angles in a parallelogram). Hence we obtain $\angle AHC=\angle AKC$, which means $AKHC$ is cyclic. This implies $\angle KH{C}_1=\angle KAC$. Then $\angle KAC=\angle ACN$, because $AK\parallel NC$ and $\angle ACN=\angle ABN$, as these angles are inscribed in the circumcircle of $ABC$. Therefore, $\angle KH{C}_1=\angle KB{C}_1$, which implies that the point $K$ also lies on the circle with diameter $BH$.

Denote by ${\omega }_1$ the circumcircle of $ABC$, by ${\omega }_2$ the circle with diameter $BH$, and by ${\omega }_3$ the circumcircle of $AKHC$. Then $AC$, $KH$ and $BD$ are radical axes of circles ${\omega }_1$ and ${\omega }_3$, ${\omega }_2$ and ${\omega }_3$, ${\omega }_1$ and ${\omega }_2$. As it is known, the radical axes of three circles either intersect at the same point, which is their radical center, or are parallel.

Since $AB>BC$, lines $AC$, $KH$ and $BD$ are concurrent.